根据条件添加行

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英文:

Adding rows depending on condition

问题

  1. 我需要帮忙解决一个非常简单的问题：
  3. 假设我有这个数据框：
  5.     data_new <- data.frame(section = c("1", "4", "5", "6"),
  6.                            density = c("0.2", "0.7", "0.8", "0.2"))
  7.     > data_new
  8.       section density
  9.     1       1     0.2
  10.     2       4     0.7
  11.     3       5     0.8
  12.     4       6     0.2
  14. 我需要添加行，因为整个表是基于6个部分，但只有4个部分的数据。这意味着在这种情况下，我需要添加2行（部分2和3），密度为0，所以我有：
  16.     > data_desired
  17.       section density
  18.     1       1     0.2
  19.     2       4     0.7
  20.     3       5     0.8
  21.     4       6     0.2
  22.     5       2       0
  23.     6       3       0
  25. 关键是*0密度*行的组合可能会变化。在这种情况下，部分3和4是空的，但下一次可能没有部分具有0密度，或者我需要添加5个部分等等。这可能会有很大的变化，从一个部分具有数据到所有部分具有数据。
  27. 我确信有一种优雅的方法可以添加到我的管道中以添加我需要的行，这是特定情况下的。非常感谢您的帮助！
英文:

I need a little help with a very simple question:

Let's say I have this data frame:

  1. data_new &lt;- data.frame(section = c(&quot;1&quot;, &quot;4&quot;, &quot;5&quot;,&quot;6&quot;),
  2.                        density = c(&quot;0.2&quot;, &quot;0.7&quot;, &quot;0.8&quot;, &quot;0.2&quot;))
  3. &gt; data_new
  4.   section density
  5. 1       1     0.2
  6. 2       4     0.7
  7. 3       5     0.8
  8. 4       6     0.2

I need to add rows because the full table is based on 6 sections, but only have data on 4. This means that in this case I have to add 2 rows (sections 2 and 3) with density 0 so I have:

  1. &gt; data_desired
  2.   section density
  3. 1       1     0.2
  4. 2       4     0.7
  5. 3       5     0.8
  6. 4       6     0.2
  7. 5       2       0
  8. 6       3       0

The point is that the combination of 0 density rows may vary. In this case sections 3 and 4 were empty, but next time it may be that no section has density 0 or that I have to add 5 sections, etc. It can vary a lot, from 1 section with data to all sections with data.

I'm sure there is an elegant way to add to my pipe to ad the rows I need and that is case specific. Thanks a lot for your help!!

答案1

得分: 1

使用 rows_update 的另一个选项：

  1. library(dplyr)
  2. # 创建零密度的数据框
  4. n <- 6
  5. data_zero <- data.frame(section = as.character(c(1:n)),
  6.                        density = as.character(rep(0,n)))
  8. data_new <- data.frame(section = c("1", "4", "5", "6"),
  9.                        density = c("0.2", "0.7", "0.8", "0.2"))
  11. rows_update(data_zero, data_new)
  12.   section density
  13. 1       1     0.2
  14. 2       2       0
  15. 3       3       0
  16. 4       4     0.7
  17. 5       5     0.8
  18. 6       6     0.2

对于多列数据：

  1. library(dplyr)
  2. n <- 6
  3. data_zero <- data.frame(section = as.character(c(1:n)),
  4.                        density = as.character(rep(0,n)))
  6. data_new <- data.frame(section = c("1", "4", "5", "6"), density = c("0.2", "0.7", "0.8", "0.2"), potatoes = c("a", "n", "ed", "3"))
  8. rows_update(data_zero, data_new[,c('section','density')]) %>%
  9.   merge(data_new, all.x = TRUE)
  11.   section density potatoes
  12. 1       1     0.2        a
  13. 2       2       0     <NA>
  14. 3       3       0     <NA>
  15. 4       4     0.7        n
  16. 5       5     0.8       ed
  17. 6       6     0.2        3
英文:

Another option using rows_update:

  1. library(dplyr)
  2. #create zero density dataframe
  4. n&lt;-6
  5. data_zero &lt;- data.frame(section = as.character(c(1:n)),
  6.                        density = as.character(rep(0,n)))
  8. data_new &lt;- data.frame(section = c(&quot;1&quot;, &quot;4&quot;, &quot;5&quot;,&quot;6&quot;),
  9.                        density = c(&quot;0.2&quot;, &quot;0.7&quot;, &quot;0.8&quot;, &quot;0.2&quot;))
  11. rows_update(data_zero ,data_new)
  12.   section density
  13. 1       1     0.2
  14. 2       2       0
  15. 3       3       0
  16. 4       4     0.7
  17. 5       5     0.8
  18. 6       6     0.2

For multiple columns:

  1. library(dplyr)
  2. n&lt;-6
  3. data_zero &lt;- data.frame(section = as.character(c(1:n)),
  4.                        density = as.character(rep(0,n)))
  7. data_new &lt;- data.frame(section = c(&quot;1&quot;, &quot;4&quot;, &quot;5&quot;,&quot;6&quot;), density = c(&quot;0.2&quot;, &quot;0.7&quot;, &quot;0.8&quot;, &quot;0.2&quot;), potatoes = c(&quot;a&quot;,&quot;n&quot;,&quot;ed&quot;,&quot;3&quot;))
  9. rows_update(data_zero ,data_new[,c(&#39;section&#39;,&#39;density&#39;)]) %&gt;%
  10.   merge(data_new,all.x=T)
  12.   section density potatoes
  13. 1       1     0.2        a
  14. 2       2       0     &lt;NA&gt;
  15. 3       3       0     &lt;NA&gt;
  16. 4       4     0.7        n
  17. 5       5     0.8       ed
  18. 6       6     0.2        3

huangapple
  • 本文由 发表于 2023年2月10日 05:27:22
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